Find the largest Perfect Subtree in a given Binary Tree

Given a Binary Tree, the task is to find the size of largest Perfect sub-tree in the given Binary Tree.
Perfect Binary Tree – A Binary tree is Perfect Binary Tree in which all internal nodes have two children and all leaves are at the same level.

Examples:

Input: 
      1
    /   \
   2     3
 /  \   /
4    5 6
Output:
Size : 3
Inorder Traversal : 4 2 5
The following sub-tree is the maximum size Perfect sub-tree 
   2  
 /  \
1    3

Input:
         50
      /      \
   30         60
  /   \      /    \ 
 5    20   45      70
          /  \     /  \
         10   85  65  80
Output:
Size : 7
Inorder Traversal : 10 45 85 60 65 70 80


Approach: Simply traverse the tree in bottom up manner. Then on coming up in recursion from child to parent, we can pass information about sub-trees to the parent. The passed information can be used by the parent to do Prefect Tree test (for parent node) only in constant time. A left sub-tree need to tell the parent whether it is a Perfect Binary Tree or not and also need to pass max height of the Perfect Binary Tree coming from left child. Similarly, the right sub-tree also needs to pass max height of Prefect Binary Tree coming from right child.
The sub-trees need to pass the following information up the tree for finding the largest Perfect sub-tree so that we can compare the maximum height with the parent’s data to check the Perfect Binary Tree property.

  1. There is a bool variable to check whether the left child or the right child sub-tree is Perfect or not.
  2. From left and right child calls in recursion we find out if parent sub-tree if Prefect or not by following 2 cases:
    • If both left child and right child are perfect binary tree and have same heights then parent is also a Perfect Binary Tree with height plus one of its child.
    • If the above case is not true then parent cannot be perfect binary tree and simply returns max size Perfect Binary Tree coming from left or right sub-tree by comparing their heights.

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Node structure of the tree
struct node {
    int data;
    struct node* left;
    struct node* right;
};
  
// To create a new node
struct node* newNode(int data)
{
    struct node* node = (struct node*)malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
    return node;
};
  
// Structure for return type of
// function findPerfectBinaryTree
struct returnType {
  
    // To store if sub-tree is perfect or not
    bool isPerfect;
  
    // Height of the tree
    int height;
  
    // Root of biggest perfect sub-tree
    node* rootTree;
};
  
// Function to return the biggest
// perfect binary sub-tree
returnType findPerfectBinaryTree(struct node* root)
{
  
    // Declaring returnType that
    // needs to be returned
    returnType rt;
  
    // If root is NULL then it is considered as
    // perfect binary tree of height 0
    if (root == NULL) {
        rt.isPerfect = true;
        rt.height = 0;
        rt.rootTree = NULL;
        return rt;
    }
  
    // Recursive call for left and right child
    returnType lv = findPerfectBinaryTree(root->left);
    returnType rv = findPerfectBinaryTree(root->right);
  
    // If both left and right sub-trees are perfect and
    // there height is also same then sub-tree root
    // is also perfect binary subtree with height
    // plus one of its child sub-trees
    if (lv.isPerfect && rv.isPerfect && lv.height == rv.height) {
        rt.height = lv.height + 1;
        rt.isPerfect = true;
        rt.rootTree = root;
        return rt;
    }
  
    // Else this sub-tree cannot be a perfect binary tree
    // and simply return the biggest sized perfect sub-tree
    // found till now in the left or right sub-trees
    rt.isPerfect = false;
    rt.height = max(lv.height, rv.height);
    rt.rootTree = (lv.height > rv.height ? lv.rootTree : rv.rootTree);
    return rt;
}
  
// Function to print the inorder traversal of the tree
void inorderPrint(node* root)
{
    if (root != NULL) {
        inorderPrint(root->left);
        cout << root->data << " ";
        inorderPrint(root->right);
    }
}
  
// Driver code
int main()
{
    // Create tree
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
  
    // Get the biggest sizes perfect binary sub-tree
    struct returnType ans = findPerfectBinaryTree(root);
  
    // Height of the found sub-tree
    int h = ans.height;
  
    cout << "Size : " << pow(2, h) - 1 << endl;
  
    // Print the inorder traversal of the found sub-tree
    cout << "Inorder Traversal : ";
    inorderPrint(ans.rootTree);
  
    return 0;
}

chevron_right


Output:

Size : 3
Inorder Traversal : 4 2 5


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.




Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.